2. Examples of continuous probability distributions: The normal and standard normal

3. The Normal Distribution X f(X)   Changing μ shifts the distribution left or right. Changing σ increases or decreases the spread.

4. The Normal Distribution: as mathematical function (pdf) Note constants:  =3.14159 e=2.71828 This is a bell shaped curve with different centers and spreads depending on  and 

5. The Normal PDF It’s a probability function, so no matter what the values of  and , must integrate to 1!

6. Normal distribution is defined by its mean and standard dev. E(X)=  = Var(X)=  2 = Standard Deviation(X)= 

7. **The beauty of the normal curve: No matter what  and  are, the area between  -  and  +  is about 68%; the area between  -2  and  +2  is about 95%; and the area between  -3  and  +3  is about 99.7%. Almost all values fall within 3 standard deviations.

8. 68-95-99.7 Rule 68% of the data 95% of the data 99.7% of the data

9. 68-95-99.7 Rule in Math terms…

10. How good is rule for real data? Check some example data: The mean of the weight of the women = 127.8 The standard deviation (SD) = 15.5

11. 127.8143.3112.3 68% of 120 =.68x120 = ~ 82 runners In fact, 79 runners fall within 1-SD (15.5 lbs) of the mean.

12. 127.896.8 95% of 120 =.95 x 120 = ~ 114 runners In fact, 115 runners fall within 2-SD’s of the mean. 158.8

13. 127.881.3 99.7% of 120 =.997 x 120 = 119.6 runners In fact, all 120 runners fall within 3-SD’s of the mean. 174.3

14. Example Suppose SAT scores roughly follows a normal distribution in the U.S. population of college-bound students (with range restricted to 200-800), and the average math SAT is 500 with a standard deviation of 50, then: 68% of students will have scores between 450 and 550 95% will be between 400 and 600 99.7% will be between 350 and 650

15. Example BUT… What if you wanted to know the math SAT score corresponding to the 90 th percentile (=90% of students are lower)? P(X≤Q) =.90  Solve for Q?….Yikes!

16. The Standard Normal (Z): “Universal Currency” The formula for the standardized normal probability density function is

17. The Standard Normal Distribution (Z ) All normal distributions can be converted into the standard normal curve by subtracting the mean and dividing by the standard deviation: Somebody calculated all the integrals for the standard normal and put them in a table! So we never have to integrate! Even better, computers now do all the integration.

18. Comparing X and Z units Z 100 2.00 200X (  = 100,  = 50) (  = 0,  = 1)

19. Example For example: What’s the probability of getting a math SAT score of 575 or less,  =500 and  =50? i.e., A score of 575 is 1.5 standard deviations above the mean Yikes! But to look up Z= 1.5 in standard normal chart (or enter into SAS)  no problem! =.9332

20. Looking up probabilities in the standard normal table What is the area to the left of Z=1.50 in a standard normal curve? Z=1.50 Area is 93.32%

21. Looking up probabilities in the standard normal table What is the area to the left of Z=1.51 in a standard normal curve? Z=1.51 Area is 93.45%

22. Probit function: the inverse  (area)= Z: gives the Z-value that goes with the probability you want For example, recall SAT math scores example. What’s the score that corresponds to the 90 th percentile? In the Table, find the Z-value that corresponds to an area of 90%...

23. 90% area corresponds to a Z score of about 1.28.

24. Probit function: the inverse Z=1.28; convert back to raw SAT score  1.28 = X – 500 =1.28 (50) X=1.28(50) + 500 = 564 (1.28 standard deviations above the mean!)

25. Practice problem If birth weights in a population are normally distributed with a mean of 109 oz and a standard deviation of 13 oz, a. What is the chance of obtaining a birth weight of 141 oz or heavier when sampling birth records at random? b. What is the chance of obtaining a birth weight of 120 or lighter?

26. Answer a. What is the chance of obtaining a birth weight of 141 oz or heavier when sampling birth records at random?

27. Area to the right of 2.46 is: 1-.9931 =.0069 or.69% Area to the left of Z=2.46 is.9931

28. Answer b. What is the chance of obtaining a birth weight of 120 or lighter?

29. Area to the left of Z=0.85 is.8023 or 80.23%.

30. Practice problem 2: DSST (a measure of cognitive function) is a normally distributed trait… Normally distributed Mean = 28 points Standard deviation = 10 points

31. Practice problem 2 a. What percent of people have values of DSST above 38? b. What percent of people have values of DSST below 8?

32. Answers a. What percent of people have values of DSST above 38? Thus, 16% of people have DSST’s above 38.

33. Answers b. What percent of people have values of DSST below 8? Thus, 2.5% of people have DSST’s below 8.

34. Review question 1 The probability that a standardized normal variable Z is positive is ____. a. 100% b. 50% c. 10% d. 0%

35. Review question 2 The probability that Z is between -2 and -1 is _____. a. 50% b. 34% c. 25.5% d. 13.5%

36. Review question 3 The probability that Z values are larger than _____ is 0.6985. a. Z=1 b. Z=0 c. Z=-.5 d. Z=+.5

37. Review question 4 27% of Z values are smaller than ____. a. Z=0 b. Z=1 c. Z=-.6 d. Z=+.6

38. Are my data “normal”? Not all continuous random variables are normally distributed!! It is important to evaluate how well the data are approximated by a normal distribution

39. Are my data normally distributed? 1.Look at the histogram! Does it appear bell shaped? 2.Compute descriptive summary measures—are mean, median, and mode similar? 3.Do 2/3 of observations lie within 1 std dev of the mean? Do 95% of observations lie within 2 std dev of the mean? 4.Look at a normal probability plot—is it approximately linear? 5.Run tests of normality (such as Kolmogorov- Smirnov). But, be cautious, highly influenced by sample size!

40. Data from our class… Median = 8 Mean = 8.8 Mode = 0 SD = 8.3 Range = 0 to 32 (= 4  )

41. Data from our class… Median = 45 Mean = 41 Mode = 6 SD = 23 Range = 0 to 83 (~ 3.5  

42. Data from our class… Median = 4 Mean = 3.7 Mode = 4 SD = 1.8 Range = 0.5 to 7 (~ 3.5  

43. Data from our class… Median = 18 Mean = 20 Mode = 20 SD = 16 Range = 2 to 70 (~4  

44. Data from our class… 8.8 +/- 8.3 = 0.5 – 17.1 0.517.1

45. Data from our class… 8.8 +/- 2*8.3 = 0 – 25.4

46. Data from our class… 8.8 +/- 3*8.3 = 0 – 33.7

47. Data from our class… 41 +/- 23 = 18 – 64 18 64

48. Data from our class… 41 +/- 2*23 = 0 – 87 087

49. Data from our class… 41 +/- 3*23 = 0– 100 0100

50. Data from our class… 3.7 +/- 1.8= 1.9 – 5.5 1.95.5

51. Data from our class… 3.7 +/- 2*1.8= 0.1 – 7.3 0.17.3

52. Data from our class… 3.7 +/- 3*1.8= 0 – 9.1 09.1

53. Data from our class… 20 +/- 16 = 4– 36 4 36

54. Data from our class… 20 +/- 2*16 = 0– 52 0 52

55. Data from our class… 20 +/- 3*16 = 0– 68 0 68 Outlier!

56. The Normal Probability Plot Normal probability plot Order the data. Find corresponding standardized normal quantile values: Plot the observed data values against normal quantile values. Evaluate the plot for evidence of linearity.

57. Normal probability plot coffee… Right-Skewed! (concave up)

58. Normal probability plot love of writing… A wiggly line!

59. Norm prob. plot Exercise… Mostly a straight line!

60. Norm prob. plot Wake up time Right-Skewed! (concave up)

61. Formal tests for normality Results: Coffee: Moderate evidence of non-normality (p=.008 to p=.11) Writing love: No evidence of non-normality (all p>.15) Exercise: No evidence of non-normality (all p>.15) Homework: Strong evidence of non-normality (all p<.01)

62. Review question 5 Which of the following does NOT support the conclusion that your data are normally distributed: a. The histogram is bell-shaped. b. The normal probability plot is approximately a straight line. c. The mean and the median are far apart. d. Formal tests of normality (with fancy Russian names) yield high p-values.

63. Normal approximation to the binomial When you have a binomial distribution where the expected value is greater than 5 (np>5), then the binomial starts to look like a normal distribution  Recall: What is the probability of being a smoker among a group of cases with lung cancer is.6, what’s the probability that in a group of 8 cases you have less than 2 smokers?

64. Normal approximation to the binomial When you have a binomial distribution where n is large and p isn’t too small (rule of thumb: mean>5), then the binomial starts to look like a normal distribution  Recall: smoking example… 145236 7 8 0.27 Starting to have a normal shape even with fairly small n. You can imagine that if n got larger, the bars would get thinner and thinner and this would look more and more like a continuous function, with a bell curve shape. Here np=4.8.

65. Normal approximation to binomial 145236 7 8 0.27 What is the probability of fewer than 2 smokers? Normal approximation probability:  =4.8  =1.39 Exact binomial probability (from before) =.00065 +.008 =.00865 P(Z<2)=.022

66. A little off, but in the right ballpark… we could also use the value to the left of 1.5 (as we really wanted to know less than but not including 2; called the “continuity correction”)… P(Z≤-2.37) =.0069 A fairly good approximation of the exact probability,.00865.

67. Practice problem 1. You are performing a cohort study. If the probability of developing disease in the exposed group is.25 for the study duration, then if you sample (randomly) 500 exposed people, What’s the probability that at most 120 people develop the disease?

68. Answer P(Z<-.52)=.3015 + + + … By hand (yikes!): P(X≤120) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)+….+ P(X=120)= OR use, normal approximation:  =np=500(.25)=125 and  2 =np(1-p)=93.75;  =9.68

69. Review question 6 If you flip a coin 1600 times, what is the approximate probability that you will get fewer than 860 heads? a. 25% b. 2.5% c. 0.5% d. 0.005%

70. Review Problem 7 Which of the following about the normal distribution is NOT true? a. Theoretically, the mean, median, and mode are the same. b. About 2/3 of the observations fall within 1 standard deviation from the mean. c. It is a discrete probability distribution. d. Its parameters are the mean, , and standard deviation, .

71. Proportions… The binomial distribution forms the basis of statistics for proportions. A proportion is just a binomial count divided by n. For example, if we sample 200 cases and find 60 smokers, X=60 but the observed proportion=.30. Statistics for proportions are similar to binomial counts, but differ by a factor of n.

72. Stats for proportions For binomial: For proportion: P-hat stands for “sample proportion.” Differs by a factor of n.

73. It all comes back to Z… Statistics for proportions are based on a normal distribution, because the binomial can be approximated as normal if np>5

74. Homework Problem Set 3 Reading: Vickers 10-15 Journal article/article review sheet